Question

How do you find the integral of `int (1 + cos x)^2 dx`?

Answer 1

`int(1+cosx)^2dx=1/4(6x+8sinx+sin2x)+"c"`

Explanation:

First expand the integrand using the perfect square formula

`int(1+cos^2x) dx=int1+2cosx+cos^2xdx`

Then use the reduction formula on `cos^2x` to get it into an integrable form

`int1+2cosx+cos^2xdx=int1+2cosx+1/2+1/2cos2xdx=int3/2+2cosx+1/2cos2xdx`

Integrate each term using the power rule or standard integrals

`int3/2+2cosx+1/2cos2x=3/2x+2sinx+1/4sin2x+"c"=1/4(6x+8sinx+sin2x)+"c"`

Answer 2

`int(1+cos x)^2 dx`
=`int(1+2cos x +cos^2 x) dx`
=`int(1+2cos x +(1/2cos 2x - 1/2) dx` *
=`int(1/2 + 2cos x + 1/2cos 2x) dx`
=`1/2x +2sin x +1/4sin 2x +c`

*`1 + 2cos^2 x = cos 2x`
`:. cos^2 x = 1/2cos 2x -1/2`